Difference between revisions of "2016 AMC 8 Problems/Problem 17"

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17. An ATM password at Fred's Bank is composed of four digits from <math>0</math> to <math>9</math>, with repeated digits allowable. If no password may begin with the sequence <math>9,1,1,</math> then how many passwords are possible?
 
17. An ATM password at Fred's Bank is composed of four digits from <math>0</math> to <math>9</math>, with repeated digits allowable. If no password may begin with the sequence <math>9,1,1,</math> then how many passwords are possible?
 
<math>(A)\mbox{ }30\mbox{          }(B)\mbox{ }7290\mbox{          }(C)\mbox{ }9000\mbox{          }(D)\mbox{ }9990\mbox{          }(E)\mbox{ }9999\mbox{          }</math>
 
<math>(A)\mbox{ }30\mbox{          }(B)\mbox{ }7290\mbox{          }(C)\mbox{ }9000\mbox{          }(D)\mbox{ }9990\mbox{          }(E)\mbox{ }9999\mbox{          }</math>
 
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==Solution==
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For the first three digits, there are <math>10^3-1=999</math> combinations since <math>911</math> is not allowed. For the finaly digit, any of the <math>10</math> numbers are allowed. <math>999 \cdot 10 = 9990 \rightarrow \boxed{D}</math>
 
{{AMC8 box|year=2016|num-b=16|num-a=18}}
 
{{AMC8 box|year=2016|num-b=16|num-a=18}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 12:36, 23 November 2016

17. An ATM password at Fred's Bank is composed of four digits from $0$ to $9$, with repeated digits allowable. If no password may begin with the sequence $9,1,1,$ then how many passwords are possible? $(A)\mbox{ }30\mbox{           }(B)\mbox{ }7290\mbox{           }(C)\mbox{ }9000\mbox{           }(D)\mbox{ }9990\mbox{           }(E)\mbox{ }9999\mbox{           }$

Solution

For the first three digits, there are $10^3-1=999$ combinations since $911$ is not allowed. For the finaly digit, any of the $10$ numbers are allowed. $999 \cdot 10 = 9990 \rightarrow \boxed{D}$

2016 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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